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Classical Mechanics

Damped Oscillators

Concept Quizzes

Exponential decay from damping forces
Damping and amplitude decreases
Damped Oscillators - Problem Solving

Damped Oscillators - Problem Solving

A body of mass $m = 1 \text{ kg}$ is oscillating on a spring with spring constant $k = 2 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -2v \text{ (N)}$ . Initially, the position of the body was $x(t=0) = 2 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the amplitude of the body when $t = 2 \text{ s } ?$

Assume that $\sqrt{2} = 1.5$

3\exp(-2) \text{ (m)}

3\exp(-3) \text{ (m)}

3\exp(-4) \text{ (m)}

4\exp(-2) \text{ (m)}

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by Brilliant Staff

A body of mass $m = 2 \text{ kg}$ is oscillating on a spring with spring constant $k = 4 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -4v \text{ (N)}$ . Initially, the position of the body was $x(t=0) = 5 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the velocity of the body when $t = \frac{\pi}{2} \text{ s } ?$

{ -9\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

{ -10\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

{ -15\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

{ -11\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

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A body of mass $m = 3 \text{ kg}$ is oscillating on a spring with spring constant $k = 3 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -6v \text{ (N)}.$ Initially, the position of the body was $x(t=0) = 1 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the velocity of the body when $t = 1 \text{ s } ?$

-1\exp(-1) \text{ (m/s)}

-1\exp(-2) \text{ (m/s)}

-2\exp(-1) \text{ (m/s)}

-1\exp(-3) \text{ (m/s)}

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by Brilliant Staff

A body of mass $m = 6 \text{ kg}$ is oscillating on a spring with spring constant $k = 12 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -12v \text{ (N)}.$ Initially, the position of the body was $x(t=0) = 4 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the ratio of the amplitude to the initial amplitude when $t = 5 \text{ s } ?$

\exp(-7)

\exp(-5)

\exp(-8)

\exp(-6)

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by Brilliant Staff

A body of mass $m = 3 \text{ kg}$ is oscillating on a spring with spring constant $k = 3 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -6v \text{ (N)}$ . Initially, the position of the body was $x(t=0) = 3 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the position of the body when $t = 3 \text{ s } ?$

12\exp(-3) \text{ (m)}

12\exp(-4) \text{ (m)}

12\exp(-5) \text{ (m)}

13\exp(-3) \text{ (m)}

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by Brilliant Staff